Symmetric Difference is Subset of Union of Symmetric Differences

Theorem

Let $R, S, T$ be sets.

Then:

$R \symdif S \subseteq \paren {R \symdif T} \cup \paren {S \symdif T}$

where $R \symdif S$ denotes the symmetric difference between $R$ and $S$.

Proof

From the definition of symmetric difference, we have:

$R \symdif S = \paren {R \setminus S} \cup \paren {S \setminus R}$

Then from Set Difference is Subset of Union of Differences, we have:

$R \setminus S \subseteq \paren {R \setminus T} \cup \paren {T \setminus S}$

$S \setminus R \subseteq \paren {S \setminus T} \cup \paren {T \setminus R}$

Thus:

\(\ds \paren {R \setminus S} \cup \paren {S \setminus R}\)

\(\subseteq\)

\(\ds \paren {R \setminus T} \cup \paren {T \setminus S} \cup \paren {S \setminus T} \cup \paren {T \setminus R}\)

Set Union Preserves Subsets

\(\ds \)

\(=\)

\(\ds \paren {R \setminus T} \cup \paren {T \setminus R} \cup \paren {S \setminus T} \cup \paren {T \setminus S}\)

Union is Commutative

\(\ds \)

\(=\)

\(\ds \paren {R \symdif T} \cup \paren {S \symdif T}\)

Definition of Symmetric Difference

$\blacksquare$